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Find the Jordan Canonical Form that is similar to the idempotent matrix $A$.

I know that since $A=A^2$ then $A(A-I)=0$ so the minimal polynomial is $m_A(\lambda)=\lambda(\lambda-1)$.

I also know that the largest Jordan block corresponding to $\lambda=0$ and $\lambda=1$ have the size $1 \mathrm x1$

But I don't really how to continue from here. I would really appreciate some help

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  • $\begingroup$ I believe that you have shown that $A$ is diagonalizable with eigenvalues $0$ and $1$. $\endgroup$ – angryavian Apr 20 at 19:38
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    $\begingroup$ From $A^2=A$ you can deduce that the minimal polynomial divides $\lambda(\lambda-1)$, but you do not know (from just that fact) that it equals this. It could be that the minimal polynomial is $\lambda$ (if $A=0$), or $\lambda-1$ (if $A$ is the identity). $\endgroup$ – Arturo Magidin Apr 20 at 19:43
  • $\begingroup$ The Jordan canonical form of $A$ is simple. It's a diagonal matrix with only zeros and ones (maybe only ones or only zeros). $\endgroup$ – amsmath Apr 20 at 19:57
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Though cannot conclude about the minimal polynomial as you did, you have already done most of the work. From $A^2=A$ you can deduce, as you say, that the only possible eigenvalues are $0$ and $1$. Next, as you mention, you deduce that the Jordan blocks can only be $1\times 1$ (because otherwise they wouldn't be idempotents). So the Jordan form of $A$ has $1\times 1$ blocks (so it's diagonal) and the diagonal consists of $0$ and $1$.

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